I am looking into a octave/matlab code and find the following:
deltaT = 1; % sampling period for data
......
R = rcValues(2:2:end)/1000; % convert these also
C = rcValues(3:2:end)*1000; % convert kF to F
RCfact = exp(-deltaT./(R.*C));
What does the point (.) mean in -deltaT. and R. in this mathematical expression?
Thanks
The dot operator is used to execute an operation on each element of a matrix. In your case, if deltaT and R are single elements, using the dot operator doesn't do anything. HOWEVER, if they were a matrix, then the operation would've been executed in each element of the matrix.
The operator is used with multiplication, division, and exponentiation.
For more info visit https://www.mathworks.com/matlabcentral/answers/506078-please-help-me-understand-the-use-of-dot-operator#accepted_answer_416043
Related
I am trying to take the derivative of a function that includes a scalar function of a vector and the vector itself. A simpler example of this is:
D[ A[b[t]]*b[t]/(b[t].b[t]), t]
where b[t] is a 3-vector and A[b[t]] is a scalar function. I get nonsense back out, since Mathematica isn't properly defining A[b[t]] to be a scalar.
I've tried using $Assumptions = {(b[t]) \[Element] Vectors[3, Reals], (M[b[t]] | t) \[Element] Reals} and this doesn't seem to help.
Any tips?
Edit to add more detail:
In that example case, I should get:
A/(b[t].b[t]) * (2(b[t].b'[t])b[t]/(b[t].b[t])^2 - b'[t])
- D[A,b]*1/(b[t].b[t])^(3/2) * (b[t].b'[t])b[t]
Where ' denotes a derivative with respect to t.
Mathematica gives everything correctly except the last term. The last term is instead:
-((b[t] b'[t] M'[b[t]])/b[t].b[t])
I've been trying to display in my console an exponential equation like the following one:
y(t) = a*e^t + b*e^t + c*e^t
I would write it as a string, however the coefficients a,b and c, are numbers in a vector V = [a b c]. So I was trying to concatenate the numbers with strings "e^t", but I failed to do it. I know scilab displays polynomial equations, but I don't know it is possible to display exponential one. Anyone can help?
Usually this kind of thing is done with mprintf command, which places given numerical arguments into a string with formatting instructions.
V = [3 5 -7]
mprintf("y(t) = %f*e^t + %f*e^t + %f*e^t", V)
The output is
y(t) = 3.000000*e^t + 5.000000*e^t + -7.000000*e^t
which isn't ideal, and can be improved in some ways by tweaking the formatters, but is readable regardless.
Notice we don't have to list every entry V(1), V(2), ... individually; the vector V gets "unpacked" automatically.
If you wanted to have 2D output like what we get for polynomials,
then no, this kind of thing is what Scilab does for polynomials and rational functions only, not for general expressions.
There is also prettyprint but its output is LaTeX syntax, like $1+s+s^{2}-s^{123}$. It works for a few things: polynomials, rational functions, matrices... but again, Scilab is not meant for symbolic manipulations, and does not really support symbolic expressions.
Suppose I have a matrix A and I want to obtain the following:
for i=1:m
A(i,:) = something which depends on i;
endfor
Is there a way to obtain that without the loop?
Added: Ok, I've understood I have to be more specific.
I have two matrices B and C (all the matrices we are considering have m rows).
I want to record in the i-th row of A the product of the polynomials written in the i-th rows of B and C (so using the loop I would call the conv function).
Any ideas?
I don't think it's possible to do this without a for loop as conv only accepts vector inputs, but I might be wrong. I can't see a way of using either bsxfun or arrayfun in conjunction with conv and matrix inputs. I might be wrong though... I stand to be corrected.
That is a very very general question, and not possible to answer with more details. Mainly on what i will be involved with. Suppose the following
for i = 1:m
A(i,:) += i;
endfor
It could be written with the much more efficient:
A .+ (1:m)'
Just compare:
octave> n = 1000;
octave> A = B = rand (n);
octave> tic; for i = 1:n, B(i,:) += i; endfor; toc
Elapsed time is 0.051 seconds.
octave> tic; C = A.+ (1:n)'; toc
Elapsed time is 0.01 seconds.
octave> isequal (C, B)
ans = 1
If you have a very old version of octave, you can instead do bsxfun (#plus, A, (i:m)').
However, if i on the right side of the expression will be used for indexing some other variable, then solution would be different. Maybe, the solution is cumsum, or some other of cumfoo function.
Your question is basically, "how do I vectorize code?", which is a really large subject, without telling us what you're trying to vectorize.
I'd like to write a Mathematica function that takes an expression as argument, takes the derivative of that expression, and then does something to the expression. So (as a toy example) I'd like to write
F[f_] = D[f, x] * 2
so that
F[x^2] = 4x
Instead, I get
F[x^2] = 0
Can someone point me to the relevant docs? I spent some time poking around the Mathematica reference, but didn't find anything helpful.
You've used assignment = when you mean to use delayed assignment :=. When you evaluate F[f_]=D[f,x]*2 using (non-delayed) assignment, Mathematica looks at D[f,x] and sees that f (an unassigned symbol) does not depend on x; hence, its derivative is 0. Thus, F[f_]=0 for any arguments to F, which is what it returns later.
If you want F to be evaluated only after you have specified what f_ should be, you need to use delayed assignment by replacing = with :=.
Started learning octave recently. How do I generate a matrix from another matrix by applying a function to each element?
eg:
Apply 2x+1 or 2x/(x^2+1) or 1/x+3 to a 3x5 matrix A.
The result should be a 3x5 matrix with the values now 2x+1
if A(1,1)=1 then after the operation with output matrix B then
B(1,1) = 2.1+1 = 3
My main concern is a function that uses the value of x like that of finding the inverse or something as indicated above.
regards.
You can try
B = A.*2 + 1
The operator . means application of the following operation * to each element of the matrix.
You will find a lot of documentation for Octave in the distribution package and on the Web. Even better, you can usually also use the extensive documentation on Matlab.
ADDED. For more complex operations you can use arrayfun(), e.g.
B = arrayfun(#(x) 2*x/(x^2+1), A)